Answer: Enthalpy of combustion (per mole) of 
is -2657.5 kJ
Explanation:
The chemical equation for the combustion of butane follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%288%5Ctimes%20%5CDelta%20H%5Eo_f_%7BCO_2%28g%29%7D%29%2B%2810%5Ctimes%20%5CDelta%20H%5Eo_f_%7BH_2O%28g%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7BC_4H_%7B10%7D%28g%29%7D%29%2B%284%5Ctimes%20%5CDelta%20H%5Eo_f_%7BO_2%28g%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%288%5Ctimes%20-393.5%29%2B%2810%5Ctimes%20-241.82%29%5D-%5B%282%5Ctimes%20-125.6%29%2B%284%5Ctimes%200%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-5315kJ)
Enthalpy of combustion (per mole) of is -2657.5 kJ
Answer:
decreased by a factor of 10
Explanation:
pH is defined in such a way that;
pH= −log10(H)
Where H represents the concentration of Hydronium or Hydrogen ions
Given that pH is changed from 1 to 2,
By rearranging the above formula , we get 10−pH = H
- if pH=1,H=10−1=0.1M
- if pH=2,H=10−2=0.01M
Therefore, 0.1/0.01 = 10 and 0.1 > 0.01
Hence, the concentration of hydronium ions in the solution is decreased by a factor of 10
Answer:
Hydrofluoric acid.
Explanation:
To know which of the acid is the strongest, let us determine the pka of each acid. This is illustrated below:
1. Acetic acid
Ka = 1.8x10^-5
pKa =..?
pKa = –logKa
pKa = –Log 1.8x10^-5
pKa = 4.74
2. Benzoic acid
Ka = 6.5x10^-5
pKa =..?
pKa = –logKa
pKa = –Log 6.5x10^-5
pKa = 4.18
3. Hydrofluoric acid.
Ka = 6.8x10^-4
pKa =..?
pKa = –logKa
pKa = –Log 6.8x10^-4
pKa = 3.17
4. Hypochlorous acid
Ka = 3.0x10^-8
pKa =..?
pKa = –logKa
pKa = –Log 3.0x10^-8
pKa = 7.52
Note: the smaller the pKa value, the stronger the acid.
The pka of the various acids as calculated above is given below:
Acid >>>>>>>>>>>>>>>>>> pKa
1. Acetic acid >>>>>>>>>> 4.74
2. Benzoic acid >>>>>>>> 4.18
3. Hydrofluoric acid >>>> 3.17
4. Hypochlorous acid >> 7.52
From the above illustration, we can see that hydrofluoric acid has the lowest pKa value. Therefore, hydrofluoric acid is the strongest among them.
Answer:
In the criss-cross method, the numerical value of the ion charge of the two atoms are crossed over, which becomes the subscript of the other ion. Using this technique, we will write the chemical formula of the given compounds.
Criss cross the absolute values to give Al2O3. To find the formula for magnesium oxide:- The oxidation number of Mg is +2 and oxygen is -2. Criss cross the absolute values to give Mg2O2In this example there is a common factor of 2 so divide by 2 to give MgO.
Answer:
The entropy change in the environment is 3.62x10²⁶.
Explanation:
The entropy change can be calculated using the following equation:
Where:
Q: is the energy transferred = 5.0 MJ
: is the Boltzmann constant = 1.38x10⁻²³ J/K
: is the initial temperature = 1000 K
: is the final temperature = 500 K
Hence, the entropy change is:
Therefore, the entropy change in the environment is 3.62x10²⁶.
I hope it helps you!
